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Given $A=\begin{bmatrix}1&1&b\\ 0&3&c\\ 0&0&1\end{bmatrix}$

Since $A$ is a triangular matrix, it has eigenvectors $\lambda_1=3$ and $\lambda_{2,3}=1$. So if I want to diagonalize this matrix, the algebraic multiplicity of each eigenvalue has to match the geometric multiplicity.

So for the eigenvalue $1$ we get:

$A-I=\begin{bmatrix}0&1&b\\ 0&2&c\\ 0&0&0\end{bmatrix}$

And how can I tell from this what the values need to be? I tried using symbolab and putting in some random numbers and sometimes it worked, sometimes it didn't. At first I thought that $2b=c$ was the condition, but it turned out false.

2 Answers 2

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You need $\ker (A - I)$ to have dimension 2 for $A$ to be diagonalizable, as $\lambda = 1$ is a double eigenvalue. This holds, iff $A$ has rank $1$, which is true iff $c = 2b$. So the condition is $c = 2b$.

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    That's what I thought but try https://www.symbolab.com/solver/matrix-eigenvectors-calculator entering some random numbers where b and c should be.2017-02-04
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    I'm not sure why, but even when that condition isn't satisfied it finds three eigenvectors, but when I try the diagonal matrix calculator it doesn't work so it seems the condition does hold, but it's confusing to say the least.2017-02-04
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    Perhaps that programs sucks. Try this one: http://wims.unice.fr/~wims/wims.cgi?session=KD3532A528.1&lang=en&cmd=reply&module=tool%2Flinear%2Fmatrix.en&matrix=1%2C-1%2C2%0D%0A-1%2C-1%2C4%0D%0A2%2C4%2C-11&show=invariant&show=charpoly&show=eigen&formula=A%5E2%2B3*A%2B2®ister=1 The condition's pretty simple and straightforward, as Martini wrote it.2017-02-04
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You have found the eigenvalues, now find the corresponding eigenvectors and create a matrix $P$. Now $P^{-1}AP = D$ where $D$ is the diagonal matrix with diagonal entries the eigenvalues. From this you can find $b,c$.

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    Why do you say the OP has found "the corresponding eigenvectors" if he hasn't? And even if had, it may be the matrix $\;P\;$ you mention *doesn't even exist*, and about this precisely is, imo, that the OP asks...2017-02-04